Verified Solution • 2025–26 Session moderate

QJEEM19Mafdd6: JEE Main 2019 Mathematics - Class 11 JEEM19Mafdd6 Solution: The domain of the definition of the function is

The domain of the definition of the function

f(x)=14−x2+log⁡10(x3−x)f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x) is

Multiple Choice Options

  • A (-1, 0) ∪\cup (1, 2) ∪\cup (2, ∞\infty )
  • B (-2, -1) ∪\cup (-1,0) ∪\cup (2, ∞\infty )
  • C (1, 2) ∪\cup (2, ∞\infty )
  • D (-1, 0) ∪\cup (1,2) ∪\cup (3, ∞\infty )

Step-by-Step Verified Solution

Given f(x)=14−x2+log⁡10(x3−x)f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)

Let f1(x) = 14−x2{1 \over {4 - {x^2}}} and f2(x) = log⁡10(x3−x){\log _{10}}({x^3} - x)

Here in f1(x) denominator ≠\ne 0

4 - x2 ≠\ne 0

⇒\Rightarrow x ≠\ne ±\pm 2 ...........(1)

and x3 - x > 0

⇒\Rightarrow x(x2 - 1) > 0

⇒\Rightarrow x(x + 1)(x - 1) > 0

JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Functions Question 113 English Explanation

x ∈\in (-1, 0) ∪\cup (1, ∞\infty ) ........(2)

Hence domain is intersection of (1) and (2)

x ∈\in (-1, 0) ∪\cup (1, 2) ∪\cup (2, ∞\infty )

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring